杭电(1009)FatMouse' Trade (贪心)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62841    Accepted Submission(s): 21210

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

 

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

 

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

 

Sample Input

   
   
   
   

    
    
    
    5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    13.333
31.500
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    题意（一直肥老鼠用f[i]的东西可以换取j[i]的爱吃的食物）
   
   
   
   
   
   
   
   

    
    
    
    相当于部分背包问题（可以分割）
   
   
   
   
   
   
   
   

    
    
    
    按照j[i]/f[i]的价值比从大往小排序。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std; struct food { int f; int j; }s[1005]; bool cmp(struct food a,struct food b) { return ((a.j*1.0)/a.f*1.0)>((b.j*1.0)/b.f*1.0); } int main() {int n,m; while(cin>>n>>m&&(n!=-1,m!=-1)) { int i; for(i=0;i<m;i++) {
        cin>>s[i].j>>s[i].f; } double sum=0;
    sort(s,s+m,cmp); for(i=0;i<m;i++) { if(n>=s[i].f) {
    
    
    
    
        n-=s[i].f;
        sum+=s[i].j; } else {
            sum+=n*((s[i].j*1.0)/s[i].f*1.0); break; } }
    printf("%.3lf\n",sum); } return 0; }